Question

在三角形ABC中,角A,B,C,的对边分别为a,b,c,求证：(a^2-b^2)/c^2=[sin(A-B)]/sinC

请写明必要过程。

Answer 1

(a^2-b^2)/c^2=(a+b/c)(a-b/c)
根据正弦定理：
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理，用和化为积公式：
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理：a-b/c=sin(A-B/2)/sin(A+B/2)
(这个公式叫模尔外得公式)
所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC

Answer 2

(a^2-b^2)/c^2=(a+b/c)(a-b/c)
根据
正弦定理
：
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理，用和化为积公式：
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理：a-b/c=sin(A-B/2)/sin(A+B/2)
所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC

Answer 3

由正弦定理:
a/sina=c/sinc
a/c=sina/sinc,两边同时乘以2cosb,左边分子分母同乘以c.得：
2ac*cosb/c²=2sinacosb/sinc.
由余弦定理a²+c²-b²=2ac*cosb得：
(a²+c²-b²)/c²=2sinacosb/sinc
两边同时减去1,可得:
(a²-b²)/c²=(2sinacosb-sinc)/sinc
且有2sinacosb-sinc=2sinacosb-sin(a+b)
=2sinacosb-(sinacosb+cosasinb)
=sinacosb-cosasinb
=sin(a-b)
则原式得证.

Answer 4

sin(A-B)/sinC=(sinAcosB-COSAsinB)/sinC=(acosB-bcosA)/c
cosB=（a²＋c²－b²）／2ac
cosA=（b²＋c²-a²）／2bc
(acosB-bcosA)/c=（a^2-b^2）/c^2
∴（a^2-b^2）/c^2=sin(A-B)/sinC